15193	2018-12-03 10:00:56	Mike Hadley	H mode mixer
Message	Date	From	Subject
Is there any advantage in using a diplexer between my H mode mixer and crystal filter. This is a 9.0 MHz IF system. Mike G4JXX Sent from my Xperia™ by Sony smartphone
15194	2018-12-03 16:15:22	Mike Nothdurft	Re: H mode mixer
I have the same question for the Tayloe mixer. Mike K5ESS
15195	2018-12-03 18:23:48	Tayloe, Dan (Noki...	Re: H mode mixer
For a tayloe mixer, the mixer looks like on open circuit on frequency, and a very low impedance far off frequency (ideally a short). The purpose of a diplexer is to terminate unwanted frequencies in a normal, non-frequency discriminating mixer. However, in this case the detector output looks like a short at undesired frequencies (except for odd harmonics of the fundamental), so it seems like a diplexer is not as useful. The unwanted frequencies still exist inside the tayloe mixer itself across the internal switch impedance (5 ohms for a ‘3253?) even when the output is a short caused by a high frequency AC signal across a detector capacitor. That is ideally (i.e., at most) a 11:1 voltage division (switch 5 ohms/(system 50 ohms + switch 5 ohms)) or a ~21 dB reduction for unwanted wide band signals inside the mixer. If you terminated the output with a diplexer so that 50 ohms was seen at all frequencies, then the detected voltage would drop 6 dB (sub optimal) and the multiplexer chip portion of the mixer would see the desired and undesired frequencies equally well, which seems suboptimal as well for spurious mixing products. It would seems offhand like terminating a tayloe mixer with a matched impedance would lower the performance. I think the tayloe detector would ideally be followed by a high impedance detector and the noise voltage would be set by the system impedance and the noise voltage for the first amplifier stage. Depending on how the quadrature detector was constructed (two outputs (0, 90 degrees) or four (0, 90, 180, 270) ) a 50 ohm input impedance might look like 200 ohms on each of 4 outputs or 100 ohms on each of two outputs at the “open circuit” detection frequency. Dan
15196	2018-12-03 18:33:37	Bill Carver	Re: H mode mixer termination
A switching mixer doesn't have The LO flowing in the mixer wires......although with device capacitances and coupling from wiring there will be a little. So unlike a diode mixer, one might expect less need for perfect termination of the mixer. However, the usual Q=1 LC diplexer (2 equal Ls, two equal C's, two 50 ohm resistors) is very simple, and very broadband. It terminates the mixer at harmonics of the LO, sum of signals and LO, etc. only a few parts, why omit it? But be aware a diplexer does not provide a proper termination at the i.f. frequency itself. At that frequency it's "transparent", and mismatch at the i.f. frequency will not be improved by the LC diplexer. Xtal filters generally have wild impedance variations on each side of the passband. Some constructors have fed filters directly from H-mode mixers and reported no horrible problem with IMD. But certainly there IS some impact. If you're trying to build the most bulletproof front end possible you should not only have a diplexer, you could deal with the impedance variation of the filter.. An xtal filter is narrowband, and it does NOT have a constant impedance, particularly in the few kHz each side of the passband. Some designers put a resistive pad (I've seen 6 dB) between the diplexer and the xtal filter. That helps, at the expense of higher noise figure in the IF subsystem. A second approach is to buffer the xtal filter with a low gain, low noise IF amplifier. NOT a Norton: it feeds load impedance variations back as an input impedance variation. Wes described the gate-source feedback amplifier in Chapter 6 of his book "Introduction to radio Frequency Design". See figure 6.11 and the few pages surrounding that figure. Figure 6.94 of Experimental Methods in RF Design" shows such a gate-source feedback amplifier as a post-filter amplifier, but it could also be used to between diplexer and filter. In this position it's not protected by the xtal filter. A third approach is to have two matched xtal filters connected together with hybrid coupler circuits. See figure 6.93 in "Experimental Methods in RF Design". A pair of matched filters will be difficult to ensure, an expensive, but the approach of this figure, using simple filters with just a few matched crystals, when terminated by the amplifier of figure 6.94, produces the 50 ohm load to the diplexer and mixer, and the amplifier buffers the single high performance filter from its impedance variations immediately below and above the filter passband. After all that........I know homebrewers using H-mode mixers who did not go to this extreme, and said they didn't observe horrible IMD. If you can construct the mixer and the Q=1 LC diplexer, try it into the xtal filter and measure, or at least critically evaluate IMD from nearby strong signals with and without a 6 dB pad ahead of the filter. Forgetting the difference in noise figure, do you measure/hear a difference in "trash" (IMD) from multiple strong nearby signals. At that point you can decide whether a more elaborate scheme is worthwhile or not. Bill - W7AAZ
15197	2018-12-03 19:05:21	Ashhar Farhan	Re: H mode mixer termination
Bill, A frequent problem that i noticed operating with an H mode without a frequency insensitive matching unit between the mixer and the filter is as follows : The mismatched antenna can show a horribly off impedance at the input of the h mode mixer which can in turn provide a mismatch to the crystal filter, resulting in a very wobbly bandpass. The antenna tunes the bandpass! - f
15198	2018-12-03 20:28:33	Bill Carver	Re: H mode mixer termination
Good input Fahran. My antennas are all matched so that's not something I've ever experienced, but since it's a switch that seems plausable. In other matters, when I go to the trouble to build something, I like it as good as possible and don't cut corners. So my experience is limited....a sort of tunnel vision. Bill
15199	2018-12-04 10:26:10	kb1gmx	Re: H mode mixer termination
Short form: Having worked with radios like HRO 50 and 60 plus many more with input tuning one thing is clear. If the antenna SWR changed the input tuning tracked it. Often the user never paid attention but the preselector or input tuning circuit was correcting for the mismatch! Essential is that this is not news only not always observed behavior and with modern wide banded gear rarely paid attention. So for those new to the idea. put an antenna tuner between a poor antenna (anything really) and the typical example of modern RX and listen to the result of matching the inefficient and reactive antenna to the front end. Its an interesting experiment, especially so if you insert a fixed attenuation of 3dB forcign a fairly fixed input impedance. SO: Diode mixers are sensitive to terminations (IF being most). They need to be matched and filtered for best results. Hmode is less sensitive depending on layout and design details. Seriously putting lousy input match is already a bad thing and if the load on the other side sees and reacts badly to me that is a design level miss and needs to be fixed or avoided. Hmode was developed to solve some of the DBM type of issues and gain a higher IP3. Tayloe/QSD is a low pass and usually only used for down conversion to baseband. That said the input port has sensitivity to source impednace. Being basically and RC system where part of the R is the a switch device and the source impedance that is the area to be paid attention to. However it seems to degrade more gracefully. The usual case for most DBMs and Hmode too is they have loss and its wise to follow with gain so the line of a mixer, diplexer, and amp with good input return loss are the usual to isolate the filter from the mixer. The classic layout of DBM, Diplexer, 2n5109 amp with 6DB pad feeding the filter generally meets the need. There are other solutions but they again address the same issues. Crystal filers are classic devices and they input impedance and behavior are very sensitive to changes and also their own reflections. With that said even basic filters have the same issues to be recognized and they all fall into the same category. They function best when terminated by a source that exhibits a fairly stable output or input impedance. Allison
15200	2018-12-07 12:00:39	iq_rx	Re: H mode mixer termination
Hi Farhan, That's a great observation. We generally think of mixers as translating the frequency of signals, but we might also think in terms of translating impedances in the frequency domain. An ideal lossless bidirectional mixer would simply translate whatever impedance is connected to the IF port at the intermediate frequency to the RF port at the radio frequency, and vice versa. A mixer isn't just a device to translate signals to a different frequency, it's a device that translates impedances to a different frequency. That's what you are observing with your low-loss H-mode mixer. A useful mental exercise it to ask, if I put a millivolt on the RF port of a mixer and 20uV flows into the RF port, what voltage appears at the IF port and how much current goes into the load connected to the IF port? Now change the load on the IF port and ask the same question. In my instrumentation receivers, I commonly add resistive attenuators to the signal path on either side of the mixers, so that filters and quadrature hybrids on one side of the mixers can work into termination impedances that don't vary as I change things on the other side. All of my recent direct conversion receivers with I Q mixers have gone a step further, with an isolating amplifier between the mixer RF ports and antenna port. That approach, with resistive attenuators in the signal path and RF gain before the mixers, is not what you want for ultimate two-tone third-order dynamic range, but if I already have more than enough of that, I'm willing to trade some for stellar in-channel noise-floor to clipping level dynamic range and in-channel gain and phase flatness. Nice work. Best Regards, Rick
15201	2018-12-07 18:25:45	iq_rx	Re: H mode mixer termination
That should have been 1mV and 20uA... Rick
15202	2018-12-07 22:28:41	Ashhar Farhan	Re: H mode mixer termination
rick, you asked: "A useful mental exercise it to ask, if I put a millivolt on the RF port of a mixer and 20uV flows into the RF port, what voltage appears at the IF port and how much current goes into the load connected to the IF port? Now change the load on the IF port and ask the same question." i am not sure i understad what you mean by "1mv flowing into the rf port and 20uv flows into the RF port", did you mean the output port? - f
15203	2018-12-08 14:19:00	iq_rx	Re: H mode mixer termination
Hi Farhan, If it's a 50 ohm port, when you put 1mV on it, 20uA should flow in. Asking where that current ends up and what voltage is associated with it is another way of thinking about port impedances. How much current flows into the port is a function of what resistance is connected to the output. It's a mental exercise, and can be mind boggling when you start thinking of lossless transformers and perfect switches in between the 1mV 20uA source and a load that you can vary to adjust how much current flows in the RF port. Best Regards, Rick
15204	2018-12-08 19:25:24	Ashhar Farhan	Re: H mode mixer termination
i guess our mails crossed each other.. i am not sure my mind can hold this puzzle together all by myself. the challenge is very interesting. i guess i will slow down the switching clock to dc, use the KISS mixer configuration as it is much simpler imagination. so now, with one switch closed, the switch is just a 1:1 transformer. it will pretty much provide this transformation all the way to its self resonance and down to where it loses its inductance. both are interesting edges. i am reminded of your paper/pdf on the four port coupler. i must simulate this and throw it together. probably time to VNAnalyze this. - f